Prove a Consequence of Cauchy sequence not converging to $0$ --- Confusion of the proof of the completion of a field.

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Let $K$ be a field and let $|\ \ |$ be an absolute value on $K$. I am reading about the completion of $K$ with respect to $|\ \ |$. Most of them make sense to me, but I got stuck in a technical step in the middle.

So let $(a_{n})_{n}$ be a $|\ \ |$-Cauchy sequence of elements $a_{n}\in K$. Suppose that $a_{n}$ does not converge to $0$ when $n\rightarrow\infty$. Then, the book I am reading claims the following:

Then, there must exist some $\epsilon_{1}>0$ and $N_{1}>0$ such that $$|a_{n}|\geq \epsilon_{1}\ \ \text{for all}\ \ n>N_{1}.$$ Otherwise, there will exist a subsequence of $a_{n}$ that converges to $0$, a contradiction.

I understand why it gives a contradiction --- if $(a_{n})_{n}$ is Cauchy that has a subsequence converging to $0$, then $(a_{n})_{n}$ must converge to $0$.


So the idea to get such an $\epsilon_{1}$ and $N_{1}$ may be to firstly realize that, if the Cauchy sequence $(a_{n})$ does not converge to $0$, then no subsequence of it can converge to $0$. Let $(a_{n_{k}})$ be a subsequence. Then, it cannot converge to $0$, which means that there exists $\epsilon_{2}>0$ such that for all $N>0$, there exists some $n_{k}>N$ such that $|a_{n_{k}}|>\epsilon_{2}$.

But how could I get an $N_{1}$ that works for all $n>N_{1}$? I am lost here.

Thank you so much for your help!